What is ##7 \times 7## in base ##7##?

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In base 7, the digits range from 0 to 6, making the multiplication of 7 by 7 initially confusing since 7 is not a valid digit in this base. The calculation shows that 7 times 7 equals 49 in base 10, which translates to 100 in base 7 after carrying over. The discussion raises questions about the validity of the problem as stated, with some participants suggesting it may imply using base 10 numbers converted to base 7. Additionally, there is confusion about how to represent base 7 multiplication in long form, but the consensus is that the answer to 7 times 7 in base 7 is indeed 100. Overall, the problem illustrates the complexities of working with different numeral systems and the importance of clear notation.
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Homework Statement
What is ##7 \times 7## in base ##7##?
Relevant Equations
Change radix to ##7##
In base ##7##, only the digits ##0,1,2,3,4,5,6## can be used. So ##7 \times 7 = 49## in base ##10##, but ##49## is a multiple of ##7## seven times, so at the right-most digit position of the answer, I can put down a ##0## and carry a ##7 ## to the next-left digit position. I can't put down ##7## in this digit position because it is bigger than the largest allowable digit, ##6##, so I put down a ##0## again, giving me a partial answer of _##00## at this step. Seven goes into seven ##1## time, so now I can put down ##1## in the next-left position and complete this question. Hence the answer is ##100##, or that in base ##7##, ##7 \times 7 = 100##.

My first question is how can ##7 \times 7## be a valid question if everything is to be in base ##7## (only zero to six can be used)? Or is this actually asking ##7 \times 7## in base ##10## and for the answer to be converted into base ##7##? I'm very confused because in the intermediary calculations I'm using the digit seven but calculating according to base ##7## rules.

My second question is how do I write down base ##7## multiplication in long hand on paper? Would something like this be correct:

##
\begin{array}{c}
1&7& \\
\hline
& &7 \\
& &7 \\
\hline
1& 0&0
\end{array}
##

I'm mostly confused here by how to treat the initial ##49##, but I couldn't see a logical place to write it down so I just went ahead and assumed seven goes into forty-nine seven times and wrote down ##7## at the very top.
 
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RChristenk said:
My first question is how can 7×7 be a valid question if everything is to be in base 7 (only zero to six can be used)?
It isn’t.
RChristenk said:
Or is this actually asking 7×7 in base 10 and for the answer to be converted into base 7?
We cannot know this. Only you have access to the source of the question.
 
RChristenk said:
Homework Statement: What is ##7 \times 7## in base ##7##?
Relevant Equations: Change radix to ##7##

In base ##7##, only the digits ##0,1,2,3,4,5,6## can be used. So ##7 \times 7 = 49## in base ##10##, but ##49## is a multiple of ##7## seven times, so at the right-most digit position of the answer, I can put down a ##0## and carry a ##7 ## to the next-left digit position. I can't put down ##7## in this digit position because it is bigger than the largest allowable digit, ##6##, so I put down a ##0## again, giving me a partial answer of _##00## at this step. Seven goes into seven ##1## time, so now I can put down ##1## in the next-left position and complete this question. Hence the answer is ##100##, or that in base ##7##, ##7 \times 7 = 100##.

My first question is how can ##7 \times 7## be a valid question if everything is to be in base ##7## (only zero to six can be used)? Or is this actually asking ##7 \times 7## in base ##10## and for the answer to be converted into base ##7##? I'm very confused because in the intermediary calculations I'm using the digit seven but calculating according to base ##7## rules.

My second question is how do I write down base ##7## multiplication in long hand on paper? Would something like this be correct:

##
\begin{array}{c}
1&7& \\
\hline
& &7 \\
& &7 \\
\hline
1& 0&0
\end{array}
##

I'm mostly confused here by how to treat the initial ##49##, but I couldn't see a logical place to write it down so I just went ahead and assumed seven goes into forty-nine seven times and wrote down ##7## at the very top.
Such a representation of a number ##a## to a base ##b## is defined as
$$
a=a_n\cdot b_n^n + \ldots + a_2\cdot 7^2+a_1\cdot 7^1+a_0\cdot 7^0
$$
with ##a_n,\ldots ,a_2,a_1,a_0 \in \{0,1,2,\ldots, b-1\}.##

This means that seven in base seven is written ##10=1\cdot 7 + 0\cdot 1.## Therefore seven times seven is
##
\begin{array}{c}
1\,0\,\;& \cdot & 1\,0\,\; \\
\hline
1\,0\,\;& & \\
\,0\,0\,0 & &\\
\hline
\,1\,0\,0&&
\end{array}
##
and ##100=1\cdot 7^2 +0\cdot 7^1=0\cdot 7^0=1\cdot 49+0\cdot 7 +0\cdot 1.##

The multiplication is the same, only with fewer numerals.
 
RChristenk said:
My first question is how can ##7 \times 7## be a valid question if everything is to be in base ##7##
The question is asking what is 710x710 in base 7?
Because if it were asking it in base 7 then the question would be what is 107 x 107? ("10" being 7 - in base 7)
And the answer would be 1007 - which is 4910.
 
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What is ##7 \times 7## in base 7?
It's reasonable to assume that the 7s above are decimal numbers (i.e., in base-10), so I agree with others that the answer is ##100_7##.
 
Perhaps I am too literal, but to me "##7 \times 7## in base ##7##" is ##10 \times 10##.
 
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Hill said:
Perhaps I am too literal, but to me "##7 \times 7## in base ##7##" is ##10 \times 10##.
See my post #3.
 
Mark44 said:
It's reasonable to assume that the 7s above are decimal numbers (i.e., in base-10), so I agree with others that the answer is ##100_7##.
Me too.

Perhaps a better wording of the problem is to use only words.

What is seven times seven ? Write the result in base seven.
 
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SammyS said:
Me too.

Perhaps a better wording of the problem is to use only words.

What is seven times seven ? Write the result in base seven.
7n x 7n = 4910
for any n larger than 7.

And we know n can't be 7 or less, since that is invalid.

Therefore, 7x7 is actually unambiguous.

75 x 75 = n/a
77 x 77 = n/a
79 x 79 = 4910
717 x 717 = 4910

In fact:

723 x 747 = 49 10
 
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  • #10
One could write the numbers as ##[a]_b## but it is pretty obvious from the context what is meant. If we are talking about ##7## in the base of ##7## then it is clear that ##[7]_{10}=[10]_7## is meant and that ##10\cdot 10 =49## is short for ##[10]_7\cdot [10]_7=[100]_7=[49]_{10}.##

There's no need to search through the crumbs.
 
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I think "b times b, in base b" is always "100" that's how I solved it, anyway.
 
  • #12
fresh_42 said:
One could write the numbers as ##[a]_b## but it is pretty obvious from the context what is meant. If we are talking about ##7## in the base of ##7## then it is clear that ##[7]_{10}=[10]_7## is meant and that ##10\cdot 10 =49## is short for ##[10]_7\cdot [10]_7=[100]_7=[49]_{10}.##

There's no need to search through the crumbs.
Still, if we work with classes, rather than numbers, which representative do we choose? [10]_7=[17]_7=...
Maybe I'm nitpicking?
 
  • #13
WWGD said:
Still, if we work with classes, rather than numbers, which representative do we choose? [10]_7=[17]_7=...
Maybe I'm nitpicking?
These weren't meant to be classes, only a notation that allows one to interpret the digits of ##a## according to different bases ##b##. We could write ##10_7=7_{10}## but I thought ##[10]_7=[7]_{10}## is easier to read. Rectangular brackets are not automatically equivalence classes.
 
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  • #14
fresh_42 said:
These weren't meant to be classes, only a notation that allows one to interpret the digits of ##a## according to different bases ##b##. We could write ##10_7=7_{10}## but I thought ##[10]_7=[7]_{10}## is easier to read. Rectangular brackets are not automatically equivalence classes.
Guess a classic/clear example of notation overload. Notation can only do, help, so much.
 

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